$H_{2} S O_{4} + N a O H \to N a H S O_{4} + H_{2} O$

In this reaction, equivalent weight of the acid is:

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None of the above

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Solution

The correct option is C 98
In the given equation,

Acid is $H_{2} S O_{4}$ .

molecular weight of $H_{2} S O_{4}$ =98g

Equivalent weight = $\frac{m o l e c u l a r w e i g h t}{n - f a c t o r}$

In this reaction, the no. of acidic hydrogen replaced in $H_{2} S O_{4}$ is 1,

so, n-factor = 1

so, equivalent weight becomes = $\frac{98}{1} = 98$

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